Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Deriving Regression Coefficients

  1. Nov 17, 2008 #1

    Sorry this may be an obvious one...can anyone help me with getting from the first to the second equation below? I'm particularly stuck with manipulating the terms inside the summations formulas.

    I can derive to here:

    \sum_{i=1}^{N}x_iy_i - N\bar{x}\bar{y} - \left( \sum_{i=1}^{N}x_i^2 - N\bar{x}^2\right)b_2 = 0

    But am a bit unsure of the exact manipulation to get to here:

    b_2 = \frac{\sum_{i=1}^{N}(x_i-y_i)(y_i-\bar{y})}{\sum_{i=1}^{N}(x_i-\bar{x})^2}

    Thanks in advance!
  2. jcsd
  3. Nov 17, 2008 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    You already know the standard representation for the regression on the slope (its your second expresssion in the original post). So try to work backwards to show that this latter expression is identical to the expression you can derive. Reversing this will then generate the path from the expression you can derive to the standard representation.
  4. Nov 17, 2008 #3
    Ah, yes...sorry, a bit obvious really when you think about it:

    \sum_{i=1}^{N}x_i^2 -N\bar{x}^2
    = \sum_{i=1}^{N}x_i^2 -\sum_{i=1}^N\bar{x}^2
    \sum_{i=1}^{N}\left( x_i - \bar{x}\right)^2

    And the other bit:

    \sum_{i=1}^{N}x_iy_i - N\bar{x}\bar{y} = \sum_{i=1}^{N}(x_i-\bar{x})(y_i-\bar{y})


    \sum_{i=1}^{N}(x_i-\bar{x})(y_i-\bar{y}) = \sum_{i=1}^{N}x_iy_i - N\bar{x}\bar{y}-N\bar{y}\bar{x}-N\bar{x}\bar{y}
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Deriving Regression Coefficients
  1. Fourier Coefficients (Replies: 1)